Lagrange Multiplier in multivariable calculus I want to ask when we apply Lagrange multiplier, we will get several critical points. And is it sufficient that the global extremum will exactly come from the value evaluated by these found critical points? In other words, is it possible for all the critical points found to be saddle points or only maximum but no minimum for example?  If not, could you please give me some examples?
What's more, is it possible to get only one critical points after applying Lagrange multiplier?
 A: Assume we have $f, g:\mathbb{R}^n \rightarrow \mathbb{R}$ differentiable and we want to find the maximum and minimum of $f$ on the set
$$
\mathcal{D} := \lbrace x\in \mathbb{R}^n: g(x) = 0 \rbrace.
$$
Then one statement (and one statement only) is always true. If $x^* \in \mathcal{D}$ is a local minimum/maximum of $f$ on $\mathcal{D}$, then there is some $\lambda \in \mathbb{R}$ such that
$$
\nabla f(x^*) + \lambda \nabla g(x^*) = 0.
$$
We call those points critical points.
What you usually do is decide if $\mathcal{D}$ is compact or not. If it in fact is, we know that a global maximum and minimum must exist. Of course, they are among the critical points. You can find them by evluating $f$ at the critical points. The lowest value is the minimum and the greatest is the maximum.
First example:
$f: \mathbb{R}^2 \rightarrow \mathbb{R}$, $f(x, y) = x+y$,  $g(x, y) = x^2+y^2 - 1$
Obviously, $\mathcal{D}$ is compact. The only critical points are $\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)^\top$ with $\lambda = \frac{1}{\sqrt{2}}$ and $\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right)^\top$ with $\lambda = -\frac{1}{\sqrt{2}}$. I think it is easy to decide which one is the minimum/maximum.
Second example: $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, $f(x, y)= \exp(-x^2-y^2)$, $g(x, y) = y$.
The only critical point is $(0, 0)$ with $\lambda = 0$. One can immediately see that it is a global maximum because of $\displaystyle \max_{t \in (-\infty, 0]} \exp(t) = \exp(0)$.
Third example: $f:\mathbb{R}^3 \rightarrow \mathbb{R}$, $f(x, y) = x^3 + y^3$, $g(x, y) = x-y$
The only critical point is $(0, 0)$ with $\lambda = 0$. It is a saddle point which you can see by looking at $f(t, 0) = t^3$. If $t>0$, $f(t, 0)$ is positive and if $t<0$, $f(t, 0)$ is negative.
(Of course other cases are possible)
